document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=639550>Question 639550</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the sum of 3 numbers is 79 the second number is 9 times the first nd the third number is three more then the second find the numbers </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #402807 by </B><A HREF=/tutors/aboutme.mpl?userid=josh_jordan><B>josh_jordan(263)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=josh_jordan><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=josh_jordan><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=402807\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To solve, we must set this question up as equations.  The first part of the  problem says that the sum of 3 numbers is 79.  In other words\r<BR>\n" );
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document.write( "x + y + z = 79\r<BR>\n" );
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document.write( "The second part of the problem says that the second number is 9 times the first.  In other words\r<BR>\n" );
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document.write( "y = 9x, where y is the second number, and x is the first number from the first equation we set up\r<BR>\n" );
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document.write( "The third part of the problem says that the third number is 3 more than the second number.  In other words\r<BR>\n" );
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document.write( "z = y + 3\r<BR>\n" );
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document.write( "To solve, we have to find each number one at a time.  So, in the first equation we set up   x + y + z = 79, the easiest number to find first is x.  So, we will substitute y and z in the first equation with y and z in the second and third equations.  Since y = 9x, in our first equation, we will substitute y with 9x, and since z = y + 3, and we know that y = 9x, we will substitute z in our first equation with 9x + 3.  So, our first equation now looks like\r<BR>\n" );
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document.write( "x + 9x + 9x + 3 = 79\r<BR>\n" );
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document.write( "Combining like terms on the left side of the equal sign gives us\r<BR>\n" );
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document.write( "19x + 3 = 79\r<BR>\n" );
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document.write( "Now, subtract 3 from both sides:\r<BR>\n" );
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document.write( "19x + 3 - 3 = 79 - 3\r<BR>\n" );
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document.write( "Which equals\r<BR>\n" );
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document.write( "19x = 76\r<BR>\n" );
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document.write( "Divide 19 by both sides so that we get x all by itself:\r<BR>\n" );
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document.write( "19x/19 = 76/19\r<BR>\n" );
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document.write( "Which gives us our first number:\r<BR>\n" );
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document.write( "x = 4\r<BR>\n" );
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document.write( "Now, we know that y = 9x, so replace x with 4: y = 9 x 4 = 36.  So, our second number is 36.\r<BR>\n" );
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document.write( "We know that z is 3 more than y, so replace y with 36 in our 3rd equation:  z = 36 + 3 = 39.  Our third number, therefore is 39\r<BR>\n" );
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document.write( "Final Answer:  Our three numbers are 4, 36, and 39\r<BR>\n" );
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document.write( " </SPAN>\n" );
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